Question

Let $p$ be the sum of all possible determinants of order $2$ having $0,1,2$ and $3$ as their four entries. Let $\alpha$ be the common root of the equations
$x^2+ax+[m+1]=0$
$x^2+bx+[m+4]=0$
$x^2-cx+[m+15]=0$
such that $\alpha >p$ and $a+b+c=0$.
If $m=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}$, then the value of $\alpha +p$ is
( $[.]$ denotes the greatest integer function)

Answer 1

$m=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n}\frac{r}{\sqrt{n^2+r^2}}$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n}\frac{r/n}{\sqrt{1+{\left(r/n\right)}^2}}$
$={\int }_0^2\frac{x}{\sqrt{1+x^2}}dx$
$={\left[\sqrt{1+x^2}\right]}_0^2$
$\therefore m=\sqrt{5}-1$
$\left[m\right]=[\sqrt{5}-1]=1$

$\alpha$ is the common root of the three equations. Then
${\alpha }^2+a\alpha +[m+1]=0\cdots \left(1\right)$
${\alpha }^2+b\alpha +[m+4]=0\cdots \left(2\right)$
${\alpha }^2-c\alpha +[m+15]=0\cdots \left(3\right)$

Adding Eqs. $\left(1\right)$ and $\left(2\right)$ and subtracting Eq. $\left(3\right)$, we get
${\alpha }^2+\left[m\right]-10=0$
$\Rightarrow {\alpha }^2-9=0$
$\Rightarrow \alpha =-3$ or $\alpha =3$

Now, number of determinants of order $2$ having $0,1,2,3$ as their entries is $4!=24$
Let ${\Delta }_1=\left|\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right|$ be one such determinant.
Then, there exists another determinant
${\Delta }_2=\left|\begin{array}{cc}{a}_3& a_4\\ a_1& a_2\end{array}\right|$ (obtained on interchanging $R_1$ and $R_2$)
such that ${\Delta }_1+{\Delta }_2=0$
So, sum of all the 24 determinants, $p=0$
Since, $\alpha >p\Rightarrow \alpha >0$
$\therefore \alpha =3$
Hence, $\alpha +p=3+0=3$